给定一个方阵M [r] [c],其中“ r”是一些行,而“ c”是列,使得r = c,我们必须检查“ M”是否为单位矩阵。
单位矩阵也称为大小为nxn方阵的单位矩阵,其中对角元素将仅具有整数值1,非对角元素将仅具有整数值0
像下面给定的示例一样-
$$I1 = \ begin {bmatrix} 1 \ end {bmatrix},\\ I2 = \ begin {bmatrix} 1&0 \\ 0&1 \ end {bmatrix},\\ I3 = \ begin {bmatrix} 1&0 &0 \\ 0&1&0 \\ 0&0&1 \ end {bmatrix},\\ In = \ begin {bmatrix}
1&0&0&...&0 \\
0&1&0&...&0 \\
0&0 &1&...&0 \\
。&。&。&...&。\\
。&。&。&...&。\\
0&0&0&...&1 \\
\ end {bmatrix} $$
Input: m[3][3] = { {1, 0, 0},
{0, 1, 0},
{0, 0, 1}}
Output: yes
Input: m[3][3] == { {3, 0, 1},
{6, 2, 0},
{7, 5, 3} }
Output: noStart Step 1 -> declare function for finding identity matrix int identity(int num) declare int row, col Loop For row = 0 and row < num and row++ Loop For col = 0 and col < num and col++ IF (row = col) Print 1 Else Print 0 End End Step 2 -> In main() Declare int size = 4 Call identity(size) Stop
#include<stdio.h>
int identity(int num){
int row, col;
for (row = 0; row < num; row++){
for (col = 0; col < num; col++){
if (row == col)
printf("%d ", 1);
else
printf("%d ", 0);
}
printf("\n");
}
return 0;
}
int main(){
int size = 4;
identity(size);
return 0;
}输出结果
1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1